Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{r - 6}{r^2 - 2r - 24} \times \dfrac{9r + 36}{-6r + 24} $
Solution: First factor the quadratic. $z = \dfrac{r - 6}{(r + 4)(r - 6)} \times \dfrac{9r + 36}{-6r + 24} $ Then factor out any other terms. $z = \dfrac{r - 6}{(r + 4)(r - 6)} \times \dfrac{9(r + 4)}{-6(r - 4)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (r - 6) \times 9(r + 4) } { (r + 4)(r - 6) \times -6(r - 4) } $ $z = \dfrac{ 9(r - 6)(r + 4)}{ -6(r + 4)(r - 6)(r - 4)} $ Notice that $(r - 6)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ 9(r - 6)\cancel{(r + 4)}}{ -6\cancel{(r + 4)}(r - 6)(r - 4)} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $z = \dfrac{ 9\cancel{(r - 6)}\cancel{(r + 4)}}{ -6\cancel{(r + 4)}\cancel{(r - 6)}(r - 4)} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $z = \dfrac{9}{-6(r - 4)} $ $z = \dfrac{-3}{2(r - 4)} ; \space r \neq -4 ; \space r \neq 6 $